544_Physics ProblemsTechnical Physics

544_Physics ProblemsTechnical Physics - 546 Superposition...

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546 Superposition and Standing Waves P18.65 (a) f n L T = 2 µ so = == f f L L L L 2 1 2 The frequency should be halved to get the same number of antinodes for twice the length. (b) = n n T T so = F H G I K J = + L N M O Q P T T n n n n 22 1 The tension must be ′= + L N M O Q P T n n T 1 2 (c) = f f nL nL T T so = ′′ F H G I K J T T nf L nfL 2 = F H G I K J T T 3 2 = T T 9 16 to get twice as many antinodes. P18.66 For the wire, × 00100 500 10 3 . . kg 2.00 m kg m: v T × 200 3 kg m s kg m 2 ej . v = 200 m s If it vibrates in its simplest state, d NN m 200 2 . λ : f v = 200 400 50 0 ms m Hz bg . . (a) The tuning fork can have frequencies 45 0 . Hz or 55.0 Hz . (b) If f = 45 0 . H z , vf = 45 0 4 00 180 .. s m m s . Then, Tv × = 2 2 3 180 5 00 10 162 kg m N . or if f = 55 0 z , f = × = 2 2 2 3 55 0 4 00 5 00 10 242 µλ . m k g m N af . P18.67 We look for a solution of the form
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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